Physics Lesson 15.5.4 - Solving Complex Circuits using the Kirchhoff Laws

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Welcome to our Physics lesson on Solving Complex Circuits using the Kirchhoff Laws, this is the fourth lesson of our suite of physics lessons covering the topic of Kirchhoff Laws, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Solving Complex Circuits using the Kirchhoff Laws

Complex circuits that cannot be solved using Ohm's Law, require a different strategy, which involves the application of Kirchhoff Laws. Each law generates a linear equation. As a result, a number of linear equation is obtained. This number depends on the number of principal nodes and loops in the circuit. There are equations involving the current law at each principal node and other equations involving the voltage law in each loop of the circuit. In general, the equations of currents in the principal nodes are equivalent, so usually we consider only one equation for currents.

As for voltages, we use a different equation for each loop. As a result, we obtain a system of linear equations that is solved using any of the three known methods of solving systems of linear equations (elimination, substitution or graph method).

The procedure for solving a circuit using the Kirchhoff Laws is as follows:

Step 1 - Make sure to write a clear circuit diagram on which you can label all known and unknown resistances, electromotive forces, and currents. If you are not sure about the direction of any current, you must anyway assign it a direction. This is necessary for determining the signs of potential differences. If you assign the direction of current incorrectly, it will result in a negative value. This is not a problem; you just realize that current is flowing in the opposite direction.

Step 2 - Apply the current rule in each principal node (junction). Every time you must get different equations, otherwise the equations are redundant, i.e. they repeat themselves. If there are only two opposite principal nodes in the circuit, the equation of currents is written only once, as the other is redundant.

Step 3 - Apply the loop rule for as many loops as needed (not necessary for each loop) to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine the signs of the potential differences and electromotive forces for each element.

Step 4 - Solve the system of linear equations. You need to express one of the currents in terms of the other two and substitute it in the system of equations involving voltages and electromotive forces.

Step 5 - Check the results you obtained for the currents substituting them in the equation of the current law. Also, make sure any resistance is not negative or it does not have an unreasonable value (very large or very small).

Now, let's consider a couple of examples where the Kirchhoff Laws are used.

Example 1

Calculate the currents flowing in all parts of the circuit below.

Solution 1

First, let's assign a positive direction to the current flow. It is better to choose the direction determined by the larger battery (here it is easy as both batteries produce a clockwise current) as the direction of current flow in the entire circuit.

Now, let's determine the equation for currents based on the First Kirchhoff Law (the law of currents). We have two principal nodes aside emf₂ and R₂. They are opposite nodes, so they produce the same (redundant) equations. Therefore, we consider only one node (for example the left one, i.e. the node A) to determine the equation of currents.

There are two currents entering the node (I₁ produced by emf₁ and I₂ produced by emf₂). Therefore, the current I₃ flowing through the resistor R₃ will leave the node A as shown in the figure.

Hence, the equation of currents will be

Now, let's determine the equations for the voltages flowing in each loop (we consider two loops: one that is determined from the lower branch (emf₁ ― R₁ ― node A ― emf₂ ― R₂ ― node B - emf₁) and the other from the higher branch (emf₂ ― node A ― R₃ ― node B ― R₂ ― emf₂). These equations form a system of linear equations shown below:

(The electromotive force emf₂ in the second source and the potential difference in the second resistor ΔV₂ = I₂ ∙ R₂ are taken as negative for the lower loop as they are in the anticlockwise direction if we take this loop as a reference, despite they have a positive direction if considering the upper loop. This situation is similar to when we raise the left hand in front of a plane mirror and it seems like our image in the mirror raises the right hand.)

Substituting the values (let's skip the units for now), we obtain

Since all values are even, we can simplify every term by 2, i.e.

This system contains three variables and two equations. Since the number of equations is smaller than that of variables, we must reduce the number of variables, i.e. we must write one of currents in terms of the other two. It is better to express the current I₃ in terms of I₁ and I₂ in the second equation. Thus, we obtain

Let's solve this system by the substitution method. Thus, from the first equation we have

Now, let's substitute this value of I₁ in the second equation and then solve for I₂.

The negative value means our assumption for the direction of current I₂ was wrong. The current I₂ flows in the other direction.

Substituting this value found for I₂ in any of the equations, we obtain

From the currents law, we obtain for the current I₃:

Thus, we must correct the direction of currents to fit the results. From the above values, we can conclude that only the current I₁ enters the node A, the other two currents leave the node. The final situation is shown in the figure below.

Therefore, the equation for currents must have been written as

We could obtain the same result by taking as the first loop the long loop that includes the entire frame of the circuit. The second loop is the same as before. Taking again the same equation for the currents (I₁ + I₂ = I₃), we write for voltages

Again, we simplify all terms by 2. Thus,

Substituting I₁ with I₂ + I₃ we have

Multiplying the first equation by 2 and adding it with the second equation, we obtain

Also,

and

In this way, we obtained the same results as when using the first method.

Now, let's see another example in which batteries produce currents in opposite directions.

Example 2

Calculate the currents flowing in the circuit shown in the figure.

Solution 2

Let's consider the principal node on the right. Given the position of the poles of each battery, we assume as a possible direction of currents the equation below

The figure below shows the flowing direction of the above currents. Also, we can take again the clockwise direction as positive.

Let's take as loops the two halves of the circuit. Hence, we obtain the following system of linear equations based on the Kirchhoff Voltage Law:

Substituting I₁ with I₂ + I₃ in the first equation, we obtain

Let's multiply the second equation by -6 to eliminate I₃ and then add the equations:

Substituting this value in the first equation (after simplifying all terms by 4 to make the calculations easier), we obtain

Therefore, we obtain for I₁:

All results are reasonable and positive, so our initial assumption for current was true.

You have reached the end of Physics lesson 15.5.4 Solving Complex Circuits using the Kirchhoff Laws. There are 4 lessons in this physics tutorial covering Kirchhoff Laws, you can access all the lessons from this tutorial below.

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